Find the derivative of the function. $$ f(t)=88^{t^{3}} $$ $$ f^{\prime}(t)=\square $$

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We start with the function 
$$
f(t)=88^{t^3}.
$$
Recall that if we have a function of the form 
$$
g(t)=a^{h(t)},
$$
its derivative is given by 
$$
g'(t)=a^{h(t)}\ln(a)\cdot h'(t).
$$

For our function, $$ a=88 $$ and $$ h(t)=t^3 $$. Now, we compute the derivative of $$ h(t) $$:
$$
h'(t)=\frac{d}{dt}(t^3)=3t^2.
$$

Thus, applying the formula we get:
$$
f'(t)=88^{t^3}\ln(88)\cdot 3t^2.
$$

Rearranging the factors, the derivative is:
$$
f'(t)=3t^2\ln(88)\cdot88^{t^3}.
$$
The derivative of $$ f(t) = 88^{t^3} $$ is $$ f'(t) = 3t^2 \ln(88) \cdot 88^{t^3} $$.

Find the derivative of the function. \[ f(t)=88^{t^{3}} \] \( f^{\prime}(t)=\square \)

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